The pressure of h2 required to make the potential of h2 electrode 0 in pure Water at 298k
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The pressure of H2 required to make the potential of H2 electrode 0 in pure water at 298K is 10-¹⁴ atm
Hydrogen ions get two ions and form hydrogen gas as below reaction.
2H+(aq) + 2e- ⇔H2(g)
according to Nernst's equation,
E- E0 = - 0.0591/n log[P_(H2)]/[[H+]²]
as electrode potential is zero in pure water. i.e., Enet = 0 when [H+] = 10^-7 [ we know, in pure water [H+] = [OH-] = 10^-7], n = 2
so, 0 = 0.0591/2 log[P/(10^-7)²]
⇒Log[P/(10-¹⁴) ] = 0
⇒P /10-¹⁴ = 1
⇒P = 10-¹⁴ atm
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